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2 years ago

6

A ball is dropped from mystery height and it accelerates at 8m/s/s and it takes .9 seconds to hit the ground what is the height

of the drop ?

1 answer:

geniusboy [140]2 years ago

6 0

Answer:

The height of the drop is 3.24 m.

Explanation:

As the ball is dropped from a mystery height and it takes 0.9 seconds to hit the ground with the acceleration of 8 m/s2. Then the velocity with which the ball is thrown can be determined as

$Velocity=Acceleration*Time$

As the acceleration is 8 m/s2 and time is 0.9 s. Then

$Velocity = 8*0.9=7.2 m/s$

So we know, the velocity u = 7.2 m/s , acceleration a = 8 m/s2 and time taken to reach the ground as t = 0.9 s. Thus, the height of the drop will be equal to the displacement of the ball from the top to the ground. Thus with the help of second equation of motion, we can determine the height as height will be equal to displacement (s).

$s=ut+\frac{1}{2}at^{2}$

Thus,

$s=(7.2*0)+(\frac{1}{2}*8*(0.9)^{2})=3.24 m$

Thus, the height of the drop will be 3.24 m.

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Answer:

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Explanation:

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The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m = mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100 °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m = mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20 °C = 39920 J

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Answer:

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Explanation:

Given:

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desity of water, $\rho=1\ kg.L^{-1}$

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$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

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$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

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